AP Physics C E and M 9.2: Electric Potential

Describe electric potential from charged objects and connect potential to the energy of a charge configuration in an electric field.

[Maximum number: 4]

A nonconducting rod of uniform negative linear charge density is near a sphere with charge +1.0 nC. The rod and sphere are held at rest on the y-axis, as shown in Figure 1. Equipotential lines and positions A, B, C, D, and E are labeled. Adjacent tick marks on the x-axis and on the y-axis are 0.40 m apart.

The electric potential $V_{\mathrm{P}}$ at $y_{\mathrm{P}}$ is $V_{\mathrm{P}}=-k \lambda \ln \left(\frac{y_{\mathrm{P}}}{y_{\mathrm{P}}-2 L}\right)$ .

Using integral calculus, derive the expression for $V_{\mathrm{P}}$ provided.

(i) For using an appropriate equation for determining the electric potential from a line of uniform charge
Example Responses
$V=\frac{1}{4 \pi \varepsilon_{0}} \sum \frac{q_{i}}{r_{i}} \quad$ OR $\quad V=\frac{1}{4 \pi \varepsilon_{0}} \int \frac{d q}{r} \quad$ OR $\quad \Delta V=-\int \vec{E} \cdot d \vec{r}$
For a correct determination of r, the distance between Point P and a point on the line of uniform charge
Example Responses
$V_{\mathrm{P}}=k \sum \frac{Q}{y_{\mathrm{P}}-y} \quad$ OR $\quad V_{\mathrm{P}}=k \int\left(\frac{1}{y_{\mathrm{P}}-y}\right) d q$
For a correct integral with $\lambda d y$ substituted for d q
Example Response
$V_{\mathrm{P}}=-k \lambda \int\left(\frac{1}{y_{\mathrm{P}}-y}\right) d y$
For the correct limits of integration
Example Response
$V_{\mathrm{P}}=-k \lambda \int_{0}^{2 L}\left(\frac{1}{y_{\mathrm{P}}-y}\right) d y$

Example Solutions

\begin{aligned} & V=\frac{1}{4 \pi \varepsilon_{0}} \sum \frac{q_{i}}{r_{i}} \\ & V_{\mathrm{P}}=k \int\left(\frac{1}{y_{\mathrm{P}}-y}\right) d q \\ & d q=-\lambda d y \\ & V_{\mathrm{P}}=-k \lambda \int\left(\frac{1}{y_{\mathrm{P}}-y}\right) d y \\ & V_{\mathrm{P}}=-k \lambda \int_{0}^{2 L}\left(\frac{1}{y_{\mathrm{P}}-y}\right) d y \\ & V_{\mathrm{P}}=\left.k \lambda \ln \left(y_{\mathrm{P}}-y\right)\right|_{0} ^{2 L}=-\left.k \lambda \ln \left(y_{\mathrm{P}}-y\right)\right|_{2 L} ^{0} \\ & V_{\mathrm{P}}=-k \lambda \ln \left(\frac{y_{\mathrm{P}}}{y_{\mathrm{P}}-2 L}\right) \end{aligned}

OR

\begin{aligned} & \Delta V=-\int \vec{E} \cdot d \vec{r} \\ & E(y)=\int \frac{k}{r^{2}} d q \\ & E(y)=\int_{0}^{2 L} \frac{k \lambda}{\left(y-y^{\prime}\right)^{2}} d y^{\prime}=k \lambda\left(\frac{1}{y-2 L}-\frac{1}{y}\right) \\ & V_{\mathrm{P}}=-\int_{\infty}^{y_{\mathrm{P}}} E(y) d y=-k \lambda \int_{\infty}^{y_{\mathrm{P}}}\left(\frac{1}{y-2 L}-\frac{1}{y}\right) d y \\ & V_{\mathrm{P}}=k \lambda \ln \left(\frac{y_{\mathrm{P}}}{y_{\mathrm{P}}-2 L}\right) \end{aligned}

All question bank results loaded

9.2.A—Describe the electric potential due to a configuration of charged objects question 1